Q: Kinematics — AQA A-Level Mathematics
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Q: Kinematics explained
Kinematics describes motion without considering its causes.
Read the full explanation
Position is where an object is relative to a chosen origin, often given as a coordinate or vector. Displacement is the change in position: a vector with magnitude and direction. Distance travelled is the total length of path covered, a scalar, so it is never negative. Velocity is the rate of change of displacement, a vector; speed is the magnitude of velocity, a scalar. Acceleration is the rate of change of velocity, a vector, so an object can accelerate by changing direction even at constant speed. In problems, choose a positive direction, then use signs consistently: negative displacement means motion in the opposite direction. For example, a ball thrown up and caught has zero displacement but non-zero distance travelled.
Understand, use and interpret graphs in kinematics for motion in a straight line: displacement against time and interpretation of gradient; velocity against time and interpretation of gradient and area under the graph.
Kinematics graphs describe motion along a straight line. On a displacement–time graph, the gradient at a point gives instantaneous velocity; a constant gradient means constant velocity, while a curve shows changing velocity. On a velocity–time graph, the gradient gives acceleration, and the area between the graph and the time axis gives displacement. Areas below the time axis count as negative displacement. For example, if v = 6 − 2t for 0 ≤ t ≤ 5, the gradient is −2 m s⁻², and the area from t = 0 to 3 is a triangle of area 9 m, while from t = 3 to 5 the area is −4 m, giving displacement 5 m. You must interpret signs, distinguish displacement from distance, and use gradients and areas to solve problems.
Understand, use and derive the formulae for constant acceleration for motion in a straight line; extend to 2 dimensions using vectors.
For motion with constant acceleration in a straight line, the SUVAT formulae link displacement s, initial velocity u, final velocity v, acceleration a and time t: v = u + at, s = ut + ½at², s = vt − ½at², v² = u² + 2as, and s = ½(u + v)t. They are derived from definitions a = dv/dt and v = ds/dt by integration, or from a velocity–time graph. In two dimensions, treat horizontal and vertical components separately using vectors: for example, r = r₀ + ut + ½at², v = u + at, with a = (0, −g) under gravity. Always define a positive direction, resolve vectors into components, and solve each component independently before combining them.
Use calculus in kinematics for motion in a straight line: v = dr/dt, a = dv/dt = d²r/dt², r = ∫v dt, v = ∫a dt; extend to 2 dimensions using vectors.
This topic connects calculus to kinematics. For straight-line motion, displacement r(t) gives velocity v = dr/dt and acceleration a = dv/dt = d²r/dt². Conversely, integrating acceleration gives velocity, and integrating velocity gives displacement, with constants found from initial conditions. In two dimensions, treat each component separately: r = xi + yj, v = ẋi + ẏj, a = ẍi + ÿj. For example, if a = 6ti + 2j, integrate to get v = 3t²i + 2tj + c; use v(0) to find c, then integrate again for r. Always check units and direction.
Model motion under gravity in a vertical plane using vectors; projectiles.
This topic models projectile motion in a vertical plane using vectors. Assuming no air resistance, acceleration is constant: a = -gj (taking j as vertical). Velocity and displacement are found by integrating: v = u + at and r = ut + ½at², where u is initial velocity. Resolve initial velocity into horizontal and vertical components. The horizontal component remains constant; the vertical component changes due to gravity. Solve problems such as finding time of flight, maximum height, range, or speed at a given time. For example, a ball projected at 20 m/s at 30° to the horizontal: u = 20cos30°i + 20sin30°j, then use equations to find position at any time.
Your focus
- Define position, displacement, distance travelled, velocity, speed and acceleration accurately.
- Classify each kinematic quantity as scalar or vector and justify the classification.
- Apply the definitions to solve one-dimensional motion problems using consistent signs.
Show all 15 objectives
- Calculate velocity as the gradient of a displacement–time graph, including at a point using a tangent.
- Calculate acceleration as the gradient of a velocity–time graph.
- Calculate displacement as the signed area under a velocity–time graph and distinguish it from total distance.
- Derive the constant-acceleration formulae from the definitions of velocity and acceleration.
- Select and use the appropriate formula to solve one-dimensional constant-acceleration problems.
- Extend the formulae to two dimensions by resolving vectors into components and solving each direction.
- Differentiate displacement with respect to time to obtain velocity and acceleration in one or two dimensions.
- Integrate acceleration with respect to time to obtain velocity and displacement, applying initial conditions.
- Solve kinematics problems involving variable acceleration using calculus and vectors.
- Resolve initial velocity into horizontal and vertical components for projectile motion.
- Apply constant acceleration equations to solve problems involving projectiles in a vertical plane.
- Interpret and critique the assumptions of the projectile model, including the effect of gravity.
Q: Kinematics exam tips
Marking Points
- Defines position as location relative to an origin and displacement as change in position, a vector.
- Distinguishes distance travelled (scalar path length) from displacement (vector change in position).
- Defines velocity as rate of change of displacement and speed as the magnitude of velocity.
- Defines acceleration as rate of change of velocity, including change of direction.
- Uses sign conventions correctly to represent direction in one-dimensional motion.
- Applies these definitions to interpret graphs or simple motion scenarios.
- State that the gradient of a displacement–time graph gives velocity, and calculate it using a chord or tangent as appropriate.
- State that the gradient of a velocity–time graph gives acceleration, and calculate it from the graph.
- State that the area under a velocity–time graph gives displacement, and calculate areas using triangles, trapeziums or rectangles.
- Interpret negative gradients and areas below the time axis as motion in the opposite direction or negative displacement.
- Distinguish between displacement and total distance travelled when the velocity changes sign.
- Use graph features such as intercepts, turning points and asymptotes to describe motion qualitatively.
- Select and apply the appropriate constant-acceleration formula to find an unknown displacement, velocity, acceleration or time.
- Derive the formulae from the definitions of velocity and acceleration or from a velocity–time graph.
- Resolve vector quantities into perpendicular components and apply the formulae to each component independently.
- Use vector equations r = r₀ + ut + ½at² and v = u + at for motion in two dimensions.
- Interpret the direction of motion from the signs of vector components and state answers with magnitude and direction where required.
- Model projectile motion by taking horizontal acceleration as zero and vertical acceleration as constant, usually −g.
- Correctly differentiate a given displacement vector to find velocity and acceleration, component by component.
- Integrate an acceleration vector to find velocity, including a constant vector determined from initial velocity.
- Integrate a velocity vector to find displacement, including a constant vector determined from initial position.
- Interpret scalar results in context, including direction and magnitude where required.
- Use calculus to solve problems involving variable acceleration in one or two dimensions.
- Resolve initial velocity into horizontal and vertical components using trigonometry.
- Apply constant acceleration equations separately to horizontal and vertical motion.
- Use vector equations v = u + at and r = ut + ½at² to find velocity and position at a given time.
- Determine key features such as time of flight, maximum height, range, and speed on impact.
- Interpret the model, including assumptions such as negligible air resistance and acceleration due to gravity.
Examiner Tips
- 💡State your positive direction at the start of a solution and keep to it throughout.
- 💡When asked to compare speed and velocity, mention both magnitude and direction explicitly.
- 💡Use the wording of the question: 'distance travelled' requires total path length, while 'displacement' requires a straight-line vector from start to finish.
- 💡Label axes with quantities and units before reading values, and show the triangle or trapezium you use for gradient or area.
- 💡When asked for distance travelled, split the motion at points where v = 0 and add the magnitudes of each area.
- 💡Check whether the question asks for a value at a specific time or over an interval; use a tangent for instantaneous rates and a chord for average rates.
- 💡List the known and unknown quantities with their directions before choosing a formula; this reduces algebraic errors.
- 💡For two-dimensional problems, draw a clear diagram and resolve all vectors into horizontal and vertical components.
- 💡Keep full accuracy through the working and round only the final answer to a sensible degree of accuracy.
- 💡Write down the given vector functions clearly and state what you are finding at each step.
- 💡Check your answer by differentiating back or ensuring units are consistent.
- 💡For two-dimensional problems, keep i and j components separate throughout your working.
- 💡Draw a clear diagram showing the initial velocity, angle, and coordinate axes.
- 💡State the equations you use and substitute values carefully, keeping components separate.
- 💡Check whether the question asks for a vector or scalar quantity, and give units where appropriate.
Common Mistakes
- Confusing distance and displacement: distance is always positive path length, displacement can be negative; correct by sketching the path and marking start and end positions.
- Treating speed and velocity as interchangeable: speed has no direction, velocity does; correct by stating direction when giving velocity.
- Assuming acceleration only means speeding up: acceleration includes slowing down and changing direction; correct by defining it as any change in velocity.
- Confusing the roles of gradient and area: for example, finding the area under a displacement–time graph instead of its gradient. Correction: gradient of s–t gives velocity; area under v–t gives displacement.
- Ignoring the sign of areas below the time axis, treating all area as positive distance. Correction: areas below the axis contribute negative displacement; total distance is the sum of absolute areas.
- Assuming a straight line on a displacement–time graph means the object is stationary. Correction: a straight line with non-zero gradient means constant velocity; only a horizontal line means stationary.
- Using a constant-acceleration formula when acceleration is not constant. Correction: check that acceleration is constant before using SUVAT; otherwise use calculus or graph methods.
- Mixing positive and negative directions within one calculation. Correction: choose a positive direction at the start and assign signs consistently to u, v, a and s.
- Treating vectors as scalars in two dimensions. Correction: resolve into components, solve each direction separately, then combine using Pythagoras and trigonometry if needed.
- Forgetting the constant of integration when finding velocity or displacement; always use initial conditions to evaluate it.
- Differentiating or integrating only one component of a vector; apply the operation to each component separately.
- Confusing displacement, velocity and acceleration notation; ensure you know which function is being differentiated or integrated.
- Using the same acceleration for horizontal and vertical motion; horizontal acceleration is zero, vertical acceleration is -g.
- Forgetting to resolve initial velocity into components; always start by finding u_x and u_y.
- Mixing up displacement and distance; displacement is a vector, distance is scalar.